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# Is 4^(x + 3) < 64 ? (1) 4^(x + 1) < 4 (2) 4^(x + 2) > 4

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Is 4^(x + 3) < 64 ? (1) 4^(x + 1) < 4 (2) 4^(x + 2) > 4  [#permalink]

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19 Jan 2018, 10:18
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Difficulty:

25% (medium)

Question Stats:

76% (01:15) correct 24% (01:58) wrong based on 73 sessions

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Is $$4^{(x + 3)} < 64$$ ?

(1) $$4^{(x + 1)} < 4$$

(2) $$4^{(x + 2)} > 4$$

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Re: Is 4^(x + 3) < 64 ? (1) 4^(x + 1) < 4 (2) 4^(x + 2) > 4  [#permalink]

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20 Jan 2018, 08:49
A.
The question stem can be rephrased as 4^(x+3)<4^3? So we have to find x+3<3 or x<0?

From st1 we have x+1<1. Hence x<0. Sufficient.

From st2 we have x+2>1. Hence x>-1. It gives many values. For example when x=-0.5 the question stem holds true. But when x>0 it doesn’t. Hence insufficient.

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Re: Is 4^(x + 3) < 64 ? (1) 4^(x + 1) < 4 (2) 4^(x + 2) > 4  [#permalink]

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22 Jan 2018, 14:20
Bunuel wrote:
Is $$4^{(x + 3)} < 64$$ ?

(1) $$4^{(x + 1)} < 4$$

(2) $$4^{(x + 2)} > 4$$

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

The question can be modified as follows.
$$4^{x+3} < 64$$ ?
$$⇔ 4^{x+3} < 4^3$$ ?
$$⇔ x+3 < 3$$ ?
$$⇔ x < 0$$ ?

Since we have 1 variable, "$$x$$" and 0 equations, D is most likely to be the answer. So, we should consider each of condition one by one first.

Condition 1):
$$4^{x+1} < 4$$
$$⇔ 4^{x+1} < 4^1$$
$$⇔ x+1 < 1$$
$$⇔ x < 0$$
The condition 1) is sufficient.

Condition 2):
$$4^{x+2} > 4$$
$$⇔ 4^{x+2} > 4^1$$
$$⇔ x+2 > 1$$
$$⇔ x > 1$$
Since the range of the question "$$x < 0$$" does not include that of the condition 2) "$$x > 1$$", the condition 2) is not sufficient.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Re: Is 4^(x + 3) < 64 ? (1) 4^(x + 1) < 4 (2) 4^(x + 2) > 4   [#permalink] 22 Jan 2018, 14:20
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